RRB Junior Engineer Stage 1 Model Papers 8

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Three solid spheres of radius 3 cm, 4 cm, and 5 cm are melted and recasted into a solid sphere. What will be the percentage decrease in the surface area?

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Solution:

Total surface area of three solid spheres

=4×22/7×(3^2+4^2+5^2)=4×22/7×50

=628.57 cm^2

Now,
Volume of new sphere

=4/3×22/7×(3^3+4^3+5^3)

∴4/3×22/7×R^3=4/3×22/7×216

R^3=216

∴R=√^3{216}=6 cm

∴ Surface Area of new solid sphere

=4×22/7×(6)^2=4×22/7×36=452.5 cm^2

∴ Required percentage

={(628.57-452.5)}/628.57×100=28%

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